# example of Cramer’s rule

Say we want to solve the system

 $\displaystyle 3x+2y+z-2w$ $\displaystyle=$ $\displaystyle 4$ $\displaystyle 2x-y+2z-5w$ $\displaystyle=$ $\displaystyle 15$ $\displaystyle 4x+2y\phantom{+2x}-5w$ $\displaystyle=$ $\displaystyle 1$ $\displaystyle 3x\phantom{+2y}-2z-4w$ $\displaystyle=$ $\displaystyle 1.$

The associated matrix is

 $\begin{pmatrix}3&2&1&-2\\ 2&-1&2&-5\\ 4&2&0&-1\\ 3&0&-2&-4\end{pmatrix}$

whose determinant  is $\Delta=-65$. Since the determinant is non-zero, we can use Cramer’s rule. To obtain the value of the $k$-th variable, we replace the $k$-th column of the matrix above by the column vector  $\begin{pmatrix}4\\ 15\\ 1\\ 1\end{pmatrix},$

the determinant of the obtained matrix is divided by $\Delta$ and the resulting value is the wanted solution.

So

 $x=\frac{\Delta_{1}}{\Delta}=\frac{\begin{vmatrix}4&2&1&-2\\ 15&-1&2&-5\\ 1&2&0&-1\\ 1&0&-2&-4\end{vmatrix}}{-65}=\frac{-65}{-65}=1$
 $y=\frac{\Delta_{2}}{\Delta}=\frac{\begin{vmatrix}3&4&1&-2\\ 2&15&2&-5\\ 4&1&0&-1\\ 3&1&-2&-4\end{vmatrix}}{-65}=\frac{130}{-65}=2$
 $z=\frac{\Delta_{3}}{\Delta}=\frac{\begin{vmatrix}3&2&4&-2\\ 2&-1&15&-5\\ 4&2&1&1\\ 3&0&1&-4\end{vmatrix}}{-65}=\frac{-195}{-65}=3$
 $w=\frac{\Delta_{4}}{\Delta}=\frac{\begin{vmatrix}3&2&1&4\\ 2&-1&2&15\\ 4&2&0&1\\ 3&0&-2&1\end{vmatrix}}{-65}=\frac{65}{-65}=-1$
Title example of Cramer’s rule ExampleOfCramersRule 2013-03-22 12:50:33 2013-03-22 12:50:33 drini (3) drini (3) 7 drini (3) Example msc 15A15